part B M = 1 & N = 15
for first digraph
R1 = { (a,b),(b,c),(c,a)}
R2 = {(a,c),(b,a),(c,b)}
R3 = { (a,b),(b,c),(c,a)}
so for first digraph m=1 and n = 3
similarly for second digraph we will let m=1 and n =5 . So to satisfy both the digraph together we will have to take LCM of (3,5) which is 15
hence m=1 and n=15